Đáp án:
Giải thích các bước giải:
`sin (9x+\frac{\pi}{3})=sin \frac{\pi}{6}`
`⇔` \(\left[ \begin{array}{l}9x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\9x+\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{\pi}{54}+k\dfrac{2\pi}{9}\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{54}+k\dfrac{2\pi}{9}\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy .......
